Problem 6 - Thomas Calculus

Question

Give a formula \(\mathbf{F}=M(x, y) \mathbf{i}+N(x, y) \mathbf{j}\) for the vector field in the plane that has the properties that \(\mathbf{F}=0\) at \((0,0)\) and that at any other point \((a, b), \mathbf{F}\) is tangent to the circle \(x^{2}+y^{2}=a^{2}+b^{2}\) and points in the clockwise direction with magnitude \(|\mathbf{F}|=\) \(\sqrt{a^{2}+b^{2}}\)

Step-by-Step Solution

Answer

The vector field is \(\mathbf{F}(x, y) = -y \mathbf{i} + x \mathbf{j}\).

Step 1: Understand the Circle Properties

The problem states that the vector field \( \mathbf{F} \) is tangent to the circle \( x^2 + y^2 = a^2 + b^2 \) at any point \((a, b)\), with vectors pointing in a clockwise direction. Essentially, this means that the vector field at \((x, y)\) should be perpendicular to the radial vector \((x, y)\) and has its direction in the orientation given by the trigonometric circle but in a clockwise manner. Consequently, the vector \(\mathbf{F}\) should have a direction given by \((-y, x)\).

Step 2: Define the Magnitude

At any point \((a, b)\) other than \((0,0)\), the magnitude of the vector field \(|\mathbf{F}|\) should correspond to \(\sqrt{a^2 + b^2}\), which matches the radius of the circle through \((a, b)\). This magnitude ensures that the length of \(\mathbf{F}\) at each point correctly scales with the given circle's properties.

Step 3: Construct the Vector Field Formula

Given that the vector is perpendicular to \((x, y)\) (the radial vector) and has the clockwise direction \((-y, x)\), the vector field can be expressed as \(\mathbf{F}(x, y) = |\mathbf{F}| \frac{(-y, x)}{\sqrt{a^2 + b^2}}\). Substituting the magnitude, the formula becomes \(\mathbf{F}(x, y) = \sqrt{x^2 + y^2}(-y, x)\), which simplifies to \(\mathbf{F}(x, y) = -y\mathbf{i} + x\mathbf{j} \).

Step 4: Ensure \(\mathbf{F} = 0\) at \((0,0)\)

It is already specified that at \((0,0)\), the vector field must be \(\mathbf{F} = 0\). Inputting \((0,0)\) into the derived formula, we get \(-0 \mathbf{i} + 0 \mathbf{j} = \mathbf{0}\), confirming this property holds true.

Key Concepts

Tangential Vectors

A key aspect of vector fields is understanding how tangential vectors work. In this case, vectors are tangent to a circle, meaning they touch the curve at a point without crossing it. If you imagine driving around a circular track, your car always moves in a direction tangent to the track at any given moment. This is why at any point \((x, y)\), the vector field is oriented perpendicular to the radial vector from the origin to the point.
To find the tangential direction for a given circle, there's a handy trick: switch the component positions and change one sign. For \((x, y)\), this tangential vector is written as \((-y, x)\). The reason for the perpendicular direction is quite simple. If a vector is \((x, y)\), a perpendicular vector must be such that their dot product is zero. And indeed, \(x\cdot (-y) + y\cdot x = 0\). This ensures the tangential vector smoothly follows the circle's edge, providing the direction for our field.

Circle Equations

The circle equation \(x^2 + y^2 = a^2 + b^2\) describes a set of points equidistant from a center. Here, the center is not at the origin but is instead moved by \(a\) and \(b\). This means any point \((a, b)\) lies on a circle centered at the origin with a radius equal to the distance to \((a, b)\) itself.
Circle equations are essential because they outline the field's structure. The vector field depicted aims to be tangent at these points, following the circle contours smoothly. You can imagine these circles as layers where each layer determines the vector's magnitude and direction. In our math equation, we deal with standard form \(x^2 + y^2 = r^2\) where \(r\) is the radius, and the vector field is tailored around these perfect circular boundaries.

Clockwise Direction

The clockwise direction for vector fields alludes to the fact that these vectors point mediating in a direction mirroring a clock's hands. Unlike the standard counterclockwise direction most mathematical conventions embrace (think about positive angles on the unit circle), the vectors point 'backwards' in this situation.
This alignment is achieved by swapping the positions of vector coordinates and flipping the sign of one component, \((-y, x)\). This switch gives a motion that feels 'backwards' or clockwise. You can remember this by visualizing an arrow moving around the circle to the right — exactly the path you'd see your watch hands simulate. This mirroring of movement is crucial for ensuring the field behaves as the problem describes.

Magnitude Calculation

Magnitude in a vector field translates to the 'length' of each vector at every point — its impact or strength. For our case, the magnitude at any point \((a, b)\) is calculated to be \(|\mathbf{F}| = \sqrt{a^2 + b^2}\). This magnitude represents the distance from the origin. Essentially, it tells how 'strong' or 'long' the vector is as it traces the circle.
Calculating magnitudes aligns with using the circle's radius, ensuring every vector in the field scales correctly around the circumference. For example, if a point is further from the center, the vector lengthens, faithfully representing how real-world forces like gravity decrease with distance. Magnitudes incorporate this scaling factor, portraying the vector field’s behavior accurately and consistently.